2NO2(g) + O3(g) → N2O5(g) + O2(g) ∆H˚ = -198 kJ ∆S˚ = -168 J·K-1 Ozone reacts with nitrogen dioxide — Thermodynamics and Kinetics Chemistry Question

∆Hrxn o = ∆Hf o (N₂O₅) + ∆Hf o (O₂) - [2∆Hf o (NO₂) + ∆Hf o (O₃)]
−198 kJ = 11 kJ + 0 - [2∆Hf o (NO₂) + 143 kJ]
so, ∆Hf o (NO₂) = 33 kJ

∆So < 0 because 3 moles of gas are converted into 2 moles of gas.

∆Go = ∆Ho – T∆So = -198000 J – 298 K(-168 J/K) = -148 kJ

Spontaneity will decrease as temperature increases because the entropy change of the process is negative. This makes the contribution of the T∆So term to the free energy positive. As temperature rises, this term contributes more to the free energy and because it is positive, the spontaneity must decrease.

Looking at the table we see from comparing experiments 1&2 that when [NO₂] increases by a factor of 1.5 the rate of reaction also increases by 1.5, so it is first order in [NO₂]. From experiments 2&3 we see that an increase in the [O₃] by a factor of 2.0 increases the rate of reaction by 2.0 so the reaction is also first order in [O₃].

The rate law is, rate = k[NO₂][O₃]
4.8×10-8 M s^-1 = k(0.0015 M)(0.0025 M)
so k = 0.0128 L mol^-1 s^-1

Mechanism #1 would suggest a rate law of rate = k[NO₂]2 and mechanism #2 would suggest a rate law of, rate = k[NO₂]2[O₃]/[O₂]. Neither of these mechanisms is consistent with the observed rate law.

The rate of the reaction would be expected to increase with an increase in temperature. Collision rate increases as temperature increases and a larger fraction of collisions have the energy necessary for reaction (Ea).